Are there inner and outer approximation theorems for arbitrary measures? The outer approximation theorem states that if $E$ is a Lebesgue measurable, then there exists a $G_\delta$ set $G$ containing $E$ such that the Lebesgue measure of $G$ equals the Lebesgue measure of $E$.  And the inner approximation theorem states that if $E$ is a Lebesgue measurable set, then there exists a $F_\sigma$ set $F$ contained in $E$ such that the Lebesgue measure of $F$ equals the Lebesgue measure of $E$.
I'm wondering if something similar is true for arbitrary measure spaces.  Let $X$ be a measure space, let $A$ be a collection of subsets of $X$, and let $m$ be a measure defined on $\sigma(A)$, the sigma algebra generated by $A$.  Let $A_{\sigma\delta}$ denote the collection of countable intersections of countable unions of elements of $A$, and let $A_{\delta\sigma}$ denote the collection of countable unions of countable intersections of elements of $A$.  My question is, for any $E\in\sigma(A)$, does there exist an $G\in A_{\sigma\delta}$ containing $E$ such that $m(F)=m(E)$?  And does there exist a $G\in A_{\delta\sigma}$ contained in $E$ such that $m(G)=m(E)$?
If so, is it also true for the $m$-completion of $\sigma(A)$? Because that's when you'd have an analogy to the Lebesgue case; the Lebesgue sigma algebra is the completion of the Borel sigma algebra with respect to Lebesgue measure, and the Borel sigma algebra is the sigma algebra generated by open intervals.
 A: In the case where $\mathcal{A}$ is an algebra, or even a ring, I believe the answer to the first question is true. It's a direct consequence of the Caratheodory extension procedure using an outer measure. Recall that this guarantees that
$$\mu(E) = \inf\Big\{\sum_{j \in \mathbb{N}}\mu(A_j) : A_j \in \mathcal{A}, \bigcup_{j \in \mathbb{N}}A_j \supseteq E \Big\}$$ 
for all $E \in \sigma(\mathcal{A})$. Now, for each $n \in \mathbb{N}$, we can find a sequence $(A^n_j)_{j \in \mathbb{N}}$ in $\mathcal{A}$ such that $\bigcup_{j \in \mathbb{N}} A^n_j \supseteq E$ and 
$$\mu(E) \geq \sum_{j \in \mathbb{N}}\mu(A^n_j) - 1/n \geq \mu\big(\bigcup_{j \in \mathbb{N}} A^n_j \big) - 1/n. $$
Since $\bigcup_{j \in \mathbb{N}} A^n_j \supseteq E$ for all $n$, then $\bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n \supseteq E$ and
$$\mu\big( \bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n\big) \leq \mu\big(\bigcup_{j \in \mathbb{N}} A^n_j \big) \leq \mu(E) + 1/n.$$
Since, this holds for all $n$, it follows that $\mu\big( \bigcap_{n \in \mathbb{N}}\bigcup_{j \in \mathbb{N}} A_j^n\big) \leq \mu(E)$. And since the reverse inequality is immediate, we can conclude. I think that the similar inner approximation result that you ask about will hold by considering an extension procedure with an inner measure, but I leave the details to you.
